Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

The correct answer/option is 2¹/³:1.

Step-by-step Explanation:
Let us analyze the process of two identical mercury drops coalescing into a single larger drop.

Step 1: Relate the radii of the drops using the conservation of volume.
Let $R$ be the radius of each small drop of mercury, and $\mathrm{R\text{'}}$ be the radius of the newly formed single large drop.
Since the total volume remains conserved when the two drops coalesce, the volume of the large drop is equal to the sum of the volumes of the two smaller drops:

$\frac{4}{3}\pi {\mathrm{R\text{'}}}^3=2\times \left(\frac{4}{3}\pi R^3\right)$

Canceling the common term $\frac{4}{3}\pi$ from both sides, we get:

${\mathrm{R\text{'}}}^3=2R^3$

Taking the cube root on both sides gives the relation between the new radius $\mathrm{R\text{'}}$ and the initial radius $R$:

$\mathrm{R\text{'}}=2^{1/3}R$

Step 2: Express the surface energies.
Surface energy ($E$) of a spherical liquid drop is given by the product of its surface tension ($T$) and its surface area ($A=4\pi r^2$):

$E=T\times 4\pi r^2$

The total initial surface energy ($E_i$) of the two small drops before coalescing is:

$E_i=2\times \left(T\times 4\pi R^2\right)=8\pi T{R}^2$

The final surface energy ($E_f$) of the single large drop after coalescing is:

$E_f=T\times 4\pi {\mathrm{R\text{'}}}^2$

Step 3: Calculate the ratio of the total surface energies before and after the change.
The ratio of the total initial surface energy to the final surface energy is:

$\frac{E_i}{E_f}=\frac{2\times T\times 4\pi R^2}{T\times 4\pi {\mathrm{R\text{'}}}^2}=\frac{2R^2}{{\mathrm{R\text{'}}}^2}$

Substituting $\mathrm{R\text{'}}=2^{1/3}R$ into the equation:

$\frac{E_i}{E_f}=\frac{2R^2}{{\left(2^{1/3}R\right)}^2}=\frac{2R^2}{2^{2/3}{R}^2}=\frac{2}{2^{2/3}}$

Simplifying the exponent:

$\frac{E_i}{E_f}=2^{1-2/3}=2^{1/3}$

Therefore, the ratio of the surface energies before and after coalescing is $2^{1/3}:1$ (or 2¹/³:1).